Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F 1 and F 2 of the lens are 5.00 cm from the center of the lens, (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the filial image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.

Question

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Answer 1

Textbook Question

Chapter 36, Problem 36.95CP

Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F₁ and F₂ of the lens are 5.00 cm from the center of the lens, (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the filial image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.

Chapter 36, Problem 36.95CP, Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have

(a)

Expert Solution

To determine

The index of refraction of the lens material.

Answer to Problem 36.95CP

The index of refraction of the lens material is $1.99$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

Write the expression for the focal length of the lens,

$1f1=(n−1)(1R1−1R2)$

Here,

$n$ is the refractive index.

$R1$ is the initial radius of curvature.

$R2$ is the final radius of curvature.

Substitute $5.00 cm$ for $f1$ , $9.00 cm$ for $R1$ and $−11.0 cm$ for $R2$ in above expression.

$15.00 cm=(n−1)(19.00 cm−1(−11.0 cm))(n−1)=0.99n=1.99$

Conclusion:

Therefore the index of refraction of the lens material is $1.99$ .

(b)

Expert Solution

To determine

The position of the final image.

Answer to Problem 36.95CP

The position of the final image is $10.0 cm$ to the left of the lens.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The expression for the focal length is,

$f2=R2$

Here,

$R$ is the radius.

Substitute $8.00 cm$ for $R$ in above expression.

$f2=8 cm2=4 cm$

Thus the focal length is $4 cm$ .

The expression for the lens equation is,

$1p1+1q1=1f1$

Here,

$q1$ is the image distance.

$p1$ is the object distance.

$f1$ is the focal length.

Substitute $5.00 cm$ for $f1$ and $8.00 cm$ for $p1$ in above expression.

$18.00 cm+1q1=15.00 cmq1=13.33 cm$

Thus the image distance from the lens is $13.33 cm$ .

The expression for the object distance for mirror is,

$p2=d−q1$

Here,

$d$ is the distance between the lens and mirror.

Substitute $13.33 cm$ for $q1$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−13.33 cm=6.67 cm$

Thus the object distance for mirror is $6.67 cm$ .

The expression for the mirror equation is,

$1p2+1q2=1f2$

Here,

$q2$ is the image distance for mirror.

$p2$ is the object distance for mirror.

$f2$ is the focal length for mirror.

Substitute $4.00 cm$ for $f2$ and $6.67 cm$ for $p2$ in above expression.

$16.67 cm+1q2=14.00 cmq2=10.0 cm$

Thus the image distance from the mirror is $10.0 cm$ .

The expression for the object distance for lens is,

$p3=d−q2$

Substitute $10.0 cm$ for $q2$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−10.0 cm=10 cm$

Thus the object distance for lens is $10.0 cm$ .

The expression for the lens equation is,

$1p3+1q3=1f3$

Here,

$q3$ is the image distance for lens.

$p3$ is the object distance for lens.

$f3$ is the focal length for lens.

Substitute $5.00 cm$ for $f3$ and $10.0 cm$ for $p3$ in above expression.

$110.0 cm+1q3=15.00 cmq3=10.0 cm$

Conclusion:

Therefore the position of the final image is $10.0 cm$ to the left of the lens.

(c)

Expert Solution

To determine

The magnification as seen by the eye in the figure.

Answer to Problem 36.95CP

The magnification as seen by the eye in the figure is $−2.50$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The overall magnification factor is,

$M=(−q1p1)(−q2p2)(−q3p3)$

Substitute $13.33 cm$ for $q1$ , $8.0 cm$ for $p1$ , $10.0 cm$ for $q2$ , $6.67 cm$ for $p2$ , $10.0 cm$ for $q3$ and $10.0 cm$ for $p3$ in above expression.

$M=(−13.33 cm8.0 cm)(−10 cm6.67 cm)(−10 cm10 cm)=−2.50$

Conclusion:

Therefore the magnification as seen by the eye in the figure is $−2.50$ .

(d)

Expert Solution

To determine

Whether the final image is inverted or upright.

Answer to Problem 36.95CP

The final image is inverted.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

From part (c) of the question the value of the magnification factor is,

$M=−2.50$

It is known that the image will be upright when the magnification factor is positive and the image will be inverted when the magnification factor is negative.

Since the magnification factor is negative so the image will be upright.

Conclusion:

Therefore the final image is inverted.

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Answer 2

Textbook Question

Chapter 36, Problem 36.95CP

Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F₁ and F₂ of the lens are 5.00 cm from the center of the lens, (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the filial image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.

Chapter 36, Problem 36.95CP, Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have

(a)

Expert Solution

To determine

The index of refraction of the lens material.

Answer to Problem 36.95CP

The index of refraction of the lens material is $1.99$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

Write the expression for the focal length of the lens,

$1f1=(n−1)(1R1−1R2)$

Here,

$n$ is the refractive index.

$R1$ is the initial radius of curvature.

$R2$ is the final radius of curvature.

Substitute $5.00 cm$ for $f1$ , $9.00 cm$ for $R1$ and $−11.0 cm$ for $R2$ in above expression.

$15.00 cm=(n−1)(19.00 cm−1(−11.0 cm))(n−1)=0.99n=1.99$

Conclusion:

Therefore the index of refraction of the lens material is $1.99$ .

(b)

Expert Solution

To determine

The position of the final image.

Answer to Problem 36.95CP

The position of the final image is $10.0 cm$ to the left of the lens.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The expression for the focal length is,

$f2=R2$

Here,

$R$ is the radius.

Substitute $8.00 cm$ for $R$ in above expression.

$f2=8 cm2=4 cm$

Thus the focal length is $4 cm$ .

The expression for the lens equation is,

$1p1+1q1=1f1$

Here,

$q1$ is the image distance.

$p1$ is the object distance.

$f1$ is the focal length.

Substitute $5.00 cm$ for $f1$ and $8.00 cm$ for $p1$ in above expression.

$18.00 cm+1q1=15.00 cmq1=13.33 cm$

Thus the image distance from the lens is $13.33 cm$ .

The expression for the object distance for mirror is,

$p2=d−q1$

Here,

$d$ is the distance between the lens and mirror.

Substitute $13.33 cm$ for $q1$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−13.33 cm=6.67 cm$

Thus the object distance for mirror is $6.67 cm$ .

The expression for the mirror equation is,

$1p2+1q2=1f2$

Here,

$q2$ is the image distance for mirror.

$p2$ is the object distance for mirror.

$f2$ is the focal length for mirror.

Substitute $4.00 cm$ for $f2$ and $6.67 cm$ for $p2$ in above expression.

$16.67 cm+1q2=14.00 cmq2=10.0 cm$

Thus the image distance from the mirror is $10.0 cm$ .

The expression for the object distance for lens is,

$p3=d−q2$

Substitute $10.0 cm$ for $q2$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−10.0 cm=10 cm$

Thus the object distance for lens is $10.0 cm$ .

The expression for the lens equation is,

$1p3+1q3=1f3$

Here,

$q3$ is the image distance for lens.

$p3$ is the object distance for lens.

$f3$ is the focal length for lens.

Substitute $5.00 cm$ for $f3$ and $10.0 cm$ for $p3$ in above expression.

$110.0 cm+1q3=15.00 cmq3=10.0 cm$

Conclusion:

Therefore the position of the final image is $10.0 cm$ to the left of the lens.

(c)

Expert Solution

To determine

The magnification as seen by the eye in the figure.

Answer to Problem 36.95CP

The magnification as seen by the eye in the figure is $−2.50$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The overall magnification factor is,

$M=(−q1p1)(−q2p2)(−q3p3)$

Substitute $13.33 cm$ for $q1$ , $8.0 cm$ for $p1$ , $10.0 cm$ for $q2$ , $6.67 cm$ for $p2$ , $10.0 cm$ for $q3$ and $10.0 cm$ for $p3$ in above expression.

$M=(−13.33 cm8.0 cm)(−10 cm6.67 cm)(−10 cm10 cm)=−2.50$

Conclusion:

Therefore the magnification as seen by the eye in the figure is $−2.50$ .

(d)

Expert Solution

To determine

Whether the final image is inverted or upright.

Answer to Problem 36.95CP

The final image is inverted.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

From part (c) of the question the value of the magnification factor is,

$M=−2.50$

It is known that the image will be upright when the magnification factor is positive and the image will be inverted when the magnification factor is negative.

Since the magnification factor is negative so the image will be upright.

Conclusion:

Therefore the final image is inverted.

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Answer 3

Textbook Question

Chapter 36, Problem 36.95CP

Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F₁ and F₂ of the lens are 5.00 cm from the center of the lens, (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the filial image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.

Chapter 36, Problem 36.95CP, Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have

(a)

Expert Solution

To determine

The index of refraction of the lens material.

Answer to Problem 36.95CP

The index of refraction of the lens material is $1.99$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

Write the expression for the focal length of the lens,

$1f1=(n−1)(1R1−1R2)$

Here,

$n$ is the refractive index.

$R1$ is the initial radius of curvature.

$R2$ is the final radius of curvature.

Substitute $5.00 cm$ for $f1$ , $9.00 cm$ for $R1$ and $−11.0 cm$ for $R2$ in above expression.

$15.00 cm=(n−1)(19.00 cm−1(−11.0 cm))(n−1)=0.99n=1.99$

Conclusion:

Therefore the index of refraction of the lens material is $1.99$ .

(b)

Expert Solution

To determine

The position of the final image.

Answer to Problem 36.95CP

The position of the final image is $10.0 cm$ to the left of the lens.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The expression for the focal length is,

$f2=R2$

Here,

$R$ is the radius.

Substitute $8.00 cm$ for $R$ in above expression.

$f2=8 cm2=4 cm$

Thus the focal length is $4 cm$ .

The expression for the lens equation is,

$1p1+1q1=1f1$

Here,

$q1$ is the image distance.

$p1$ is the object distance.

$f1$ is the focal length.

Substitute $5.00 cm$ for $f1$ and $8.00 cm$ for $p1$ in above expression.

$18.00 cm+1q1=15.00 cmq1=13.33 cm$

Thus the image distance from the lens is $13.33 cm$ .

The expression for the object distance for mirror is,

$p2=d−q1$

Here,

$d$ is the distance between the lens and mirror.

Substitute $13.33 cm$ for $q1$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−13.33 cm=6.67 cm$

Thus the object distance for mirror is $6.67 cm$ .

The expression for the mirror equation is,

$1p2+1q2=1f2$

Here,

$q2$ is the image distance for mirror.

$p2$ is the object distance for mirror.

$f2$ is the focal length for mirror.

Substitute $4.00 cm$ for $f2$ and $6.67 cm$ for $p2$ in above expression.

$16.67 cm+1q2=14.00 cmq2=10.0 cm$

Thus the image distance from the mirror is $10.0 cm$ .

The expression for the object distance for lens is,

$p3=d−q2$

Substitute $10.0 cm$ for $q2$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−10.0 cm=10 cm$

Thus the object distance for lens is $10.0 cm$ .

The expression for the lens equation is,

$1p3+1q3=1f3$

Here,

$q3$ is the image distance for lens.

$p3$ is the object distance for lens.

$f3$ is the focal length for lens.

Substitute $5.00 cm$ for $f3$ and $10.0 cm$ for $p3$ in above expression.

$110.0 cm+1q3=15.00 cmq3=10.0 cm$

Conclusion:

Therefore the position of the final image is $10.0 cm$ to the left of the lens.

(c)

Expert Solution

To determine

The magnification as seen by the eye in the figure.

Answer to Problem 36.95CP

The magnification as seen by the eye in the figure is $−2.50$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The overall magnification factor is,

$M=(−q1p1)(−q2p2)(−q3p3)$

Substitute $13.33 cm$ for $q1$ , $8.0 cm$ for $p1$ , $10.0 cm$ for $q2$ , $6.67 cm$ for $p2$ , $10.0 cm$ for $q3$ and $10.0 cm$ for $p3$ in above expression.

$M=(−13.33 cm8.0 cm)(−10 cm6.67 cm)(−10 cm10 cm)=−2.50$

Conclusion:

Therefore the magnification as seen by the eye in the figure is $−2.50$ .

(d)

Expert Solution

To determine

Whether the final image is inverted or upright.

Answer to Problem 36.95CP

The final image is inverted.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

From part (c) of the question the value of the magnification factor is,

$M=−2.50$

It is known that the image will be upright when the magnification factor is positive and the image will be inverted when the magnification factor is negative.

Since the magnification factor is negative so the image will be upright.

Conclusion:

Therefore the final image is inverted.

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Answer 4

Textbook Question

Chapter 36, Problem 36.95CP

Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F₁ and F₂ of the lens are 5.00 cm from the center of the lens, (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the filial image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.

Chapter 36, Problem 36.95CP, Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have

(a)

Expert Solution

To determine

The index of refraction of the lens material.

Answer to Problem 36.95CP

The index of refraction of the lens material is $1.99$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

Write the expression for the focal length of the lens,

$1f1=(n−1)(1R1−1R2)$

Here,

$n$ is the refractive index.

$R1$ is the initial radius of curvature.

$R2$ is the final radius of curvature.

Substitute $5.00 cm$ for $f1$ , $9.00 cm$ for $R1$ and $−11.0 cm$ for $R2$ in above expression.

$15.00 cm=(n−1)(19.00 cm−1(−11.0 cm))(n−1)=0.99n=1.99$

Conclusion:

Therefore the index of refraction of the lens material is $1.99$ .

(b)

Expert Solution

To determine

The position of the final image.

Answer to Problem 36.95CP

The position of the final image is $10.0 cm$ to the left of the lens.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The expression for the focal length is,

$f2=R2$

Here,

$R$ is the radius.

Substitute $8.00 cm$ for $R$ in above expression.

$f2=8 cm2=4 cm$

Thus the focal length is $4 cm$ .

The expression for the lens equation is,

$1p1+1q1=1f1$

Here,

$q1$ is the image distance.

$p1$ is the object distance.

$f1$ is the focal length.

Substitute $5.00 cm$ for $f1$ and $8.00 cm$ for $p1$ in above expression.

$18.00 cm+1q1=15.00 cmq1=13.33 cm$

Thus the image distance from the lens is $13.33 cm$ .

The expression for the object distance for mirror is,

$p2=d−q1$

Here,

$d$ is the distance between the lens and mirror.

Substitute $13.33 cm$ for $q1$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−13.33 cm=6.67 cm$

Thus the object distance for mirror is $6.67 cm$ .

The expression for the mirror equation is,

$1p2+1q2=1f2$

Here,

$q2$ is the image distance for mirror.

$p2$ is the object distance for mirror.

$f2$ is the focal length for mirror.

Substitute $4.00 cm$ for $f2$ and $6.67 cm$ for $p2$ in above expression.

$16.67 cm+1q2=14.00 cmq2=10.0 cm$

Thus the image distance from the mirror is $10.0 cm$ .

The expression for the object distance for lens is,

$p3=d−q2$

Substitute $10.0 cm$ for $q2$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−10.0 cm=10 cm$

Thus the object distance for lens is $10.0 cm$ .

The expression for the lens equation is,

$1p3+1q3=1f3$

Here,

$q3$ is the image distance for lens.

$p3$ is the object distance for lens.

$f3$ is the focal length for lens.

Substitute $5.00 cm$ for $f3$ and $10.0 cm$ for $p3$ in above expression.

$110.0 cm+1q3=15.00 cmq3=10.0 cm$

Conclusion:

Therefore the position of the final image is $10.0 cm$ to the left of the lens.

(c)

Expert Solution

To determine

The magnification as seen by the eye in the figure.

Answer to Problem 36.95CP

The magnification as seen by the eye in the figure is $−2.50$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The overall magnification factor is,

$M=(−q1p1)(−q2p2)(−q3p3)$

Substitute $13.33 cm$ for $q1$ , $8.0 cm$ for $p1$ , $10.0 cm$ for $q2$ , $6.67 cm$ for $p2$ , $10.0 cm$ for $q3$ and $10.0 cm$ for $p3$ in above expression.

$M=(−13.33 cm8.0 cm)(−10 cm6.67 cm)(−10 cm10 cm)=−2.50$

Conclusion:

Therefore the magnification as seen by the eye in the figure is $−2.50$ .

(d)

Expert Solution

To determine

Whether the final image is inverted or upright.

Answer to Problem 36.95CP

The final image is inverted.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

From part (c) of the question the value of the magnification factor is,

$M=−2.50$

It is known that the image will be upright when the magnification factor is positive and the image will be inverted when the magnification factor is negative.

Since the magnification factor is negative so the image will be upright.

Conclusion:

Therefore the final image is inverted.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The index of refraction of the lens material.

Answer to Problem 36.95CP

The index of refraction of the lens material is $1.99$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

Write the expression for the focal length of the lens,

$1f1=(n−1)(1R1−1R2)$

Here,

$n$ is the refractive index.

$R1$ is the initial radius of curvature.

$R2$ is the final radius of curvature.

Substitute $5.00 cm$ for $f1$ , $9.00 cm$ for $R1$ and $−11.0 cm$ for $R2$ in above expression.

$15.00 cm=(n−1)(19.00 cm−1(−11.0 cm))(n−1)=0.99n=1.99$

Conclusion:

Therefore the index of refraction of the lens material is $1.99$ .

(b)

Expert Solution

To determine

The position of the final image.

Answer to Problem 36.95CP

The position of the final image is $10.0 cm$ to the left of the lens.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The expression for the focal length is,

$f2=R2$

Here,

$R$ is the radius.

Substitute $8.00 cm$ for $R$ in above expression.

$f2=8 cm2=4 cm$

Thus the focal length is $4 cm$ .

The expression for the lens equation is,

$1p1+1q1=1f1$

Here,

$q1$ is the image distance.

$p1$ is the object distance.

$f1$ is the focal length.

Substitute $5.00 cm$ for $f1$ and $8.00 cm$ for $p1$ in above expression.

$18.00 cm+1q1=15.00 cmq1=13.33 cm$

Thus the image distance from the lens is $13.33 cm$ .

The expression for the object distance for mirror is,

$p2=d−q1$

Here,

$d$ is the distance between the lens and mirror.

Substitute $13.33 cm$ for $q1$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−13.33 cm=6.67 cm$

Thus the object distance for mirror is $6.67 cm$ .

The expression for the mirror equation is,

$1p2+1q2=1f2$

Here,

$q2$ is the image distance for mirror.

$p2$ is the object distance for mirror.

$f2$ is the focal length for mirror.

Substitute $4.00 cm$ for $f2$ and $6.67 cm$ for $p2$ in above expression.

$16.67 cm+1q2=14.00 cmq2=10.0 cm$

Thus the image distance from the mirror is $10.0 cm$ .

The expression for the object distance for lens is,

$p3=d−q2$

Substitute $10.0 cm$ for $q2$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−10.0 cm=10 cm$

Thus the object distance for lens is $10.0 cm$ .

The expression for the lens equation is,

$1p3+1q3=1f3$

Here,

$q3$ is the image distance for lens.

$p3$ is the object distance for lens.

$f3$ is the focal length for lens.

Substitute $5.00 cm$ for $f3$ and $10.0 cm$ for $p3$ in above expression.

$110.0 cm+1q3=15.00 cmq3=10.0 cm$

Conclusion:

Therefore the position of the final image is $10.0 cm$ to the left of the lens.

(c)

Expert Solution

To determine

The magnification as seen by the eye in the figure.

Answer to Problem 36.95CP

The magnification as seen by the eye in the figure is $−2.50$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The overall magnification factor is,

$M=(−q1p1)(−q2p2)(−q3p3)$

Substitute $13.33 cm$ for $q1$ , $8.0 cm$ for $p1$ , $10.0 cm$ for $q2$ , $6.67 cm$ for $p2$ , $10.0 cm$ for $q3$ and $10.0 cm$ for $p3$ in above expression.

$M=(−13.33 cm8.0 cm)(−10 cm6.67 cm)(−10 cm10 cm)=−2.50$

Conclusion:

Therefore the magnification as seen by the eye in the figure is $−2.50$ .

(d)

Expert Solution

To determine

Whether the final image is inverted or upright.

Answer to Problem 36.95CP

The final image is inverted.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

From part (c) of the question the value of the magnification factor is,

$M=−2.50$

It is known that the image will be upright when the magnification factor is positive and the image will be inverted when the magnification factor is negative.

Since the magnification factor is negative so the image will be upright.

Conclusion:

Therefore the final image is inverted.

Answer 8

(a)

Expert Solution

To determine

The index of refraction of the lens material.

Answer to Problem 36.95CP

The index of refraction of the lens material is $1.99$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

Write the expression for the focal length of the lens,

$1f1=(n−1)(1R1−1R2)$

Here,

$n$ is the refractive index.

$R1$ is the initial radius of curvature.

$R2$ is the final radius of curvature.

Substitute $5.00 cm$ for $f1$ , $9.00 cm$ for $R1$ and $−11.0 cm$ for $R2$ in above expression.

$15.00 cm=(n−1)(19.00 cm−1(−11.0 cm))(n−1)=0.99n=1.99$

Conclusion:

Therefore the index of refraction of the lens material is $1.99$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The index of refraction of the lens material.

Answer 13

Answer to Problem 36.95CP

The index of refraction of the lens material is $1.99$ .

Answer 14

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

Write the expression for the focal length of the lens,

$1f1=(n−1)(1R1−1R2)$

Here,

$n$ is the refractive index.

$R1$ is the initial radius of curvature.

$R2$ is the final radius of curvature.

Substitute $5.00 cm$ for $f1$ , $9.00 cm$ for $R1$ and $−11.0 cm$ for $R2$ in above expression.

$15.00 cm=(n−1)(19.00 cm−1(−11.0 cm))(n−1)=0.99n=1.99$

Conclusion:

Therefore the index of refraction of the lens material is $1.99$ .

Answer 15

(b)

Expert Solution

To determine

The position of the final image.

Answer to Problem 36.95CP

The position of the final image is $10.0 cm$ to the left of the lens.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The expression for the focal length is,

$f2=R2$

Here,

$R$ is the radius.

Substitute $8.00 cm$ for $R$ in above expression.

$f2=8 cm2=4 cm$

Thus the focal length is $4 cm$ .

The expression for the lens equation is,

$1p1+1q1=1f1$

Here,

$q1$ is the image distance.

$p1$ is the object distance.

$f1$ is the focal length.

Substitute $5.00 cm$ for $f1$ and $8.00 cm$ for $p1$ in above expression.

$18.00 cm+1q1=15.00 cmq1=13.33 cm$

Thus the image distance from the lens is $13.33 cm$ .

The expression for the object distance for mirror is,

$p2=d−q1$

Here,

$d$ is the distance between the lens and mirror.

Substitute $13.33 cm$ for $q1$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−13.33 cm=6.67 cm$

Thus the object distance for mirror is $6.67 cm$ .

The expression for the mirror equation is,

$1p2+1q2=1f2$

Here,

$q2$ is the image distance for mirror.

$p2$ is the object distance for mirror.

$f2$ is the focal length for mirror.

Substitute $4.00 cm$ for $f2$ and $6.67 cm$ for $p2$ in above expression.

$16.67 cm+1q2=14.00 cmq2=10.0 cm$

Thus the image distance from the mirror is $10.0 cm$ .

The expression for the object distance for lens is,

$p3=d−q2$

Substitute $10.0 cm$ for $q2$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−10.0 cm=10 cm$

Thus the object distance for lens is $10.0 cm$ .

The expression for the lens equation is,

$1p3+1q3=1f3$

Here,

$q3$ is the image distance for lens.

$p3$ is the object distance for lens.

$f3$ is the focal length for lens.

Substitute $5.00 cm$ for $f3$ and $10.0 cm$ for $p3$ in above expression.

$110.0 cm+1q3=15.00 cmq3=10.0 cm$

Conclusion:

Therefore the position of the final image is $10.0 cm$ to the left of the lens.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The position of the final image.

Answer 20

Answer to Problem 36.95CP

The position of the final image is $10.0 cm$ to the left of the lens.

Answer 21

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The expression for the focal length is,

$f2=R2$

Here,

$R$ is the radius.

Substitute $8.00 cm$ for $R$ in above expression.

$f2=8 cm2=4 cm$

Thus the focal length is $4 cm$ .

The expression for the lens equation is,

$1p1+1q1=1f1$

Here,

$q1$ is the image distance.

$p1$ is the object distance.

$f1$ is the focal length.

Substitute $5.00 cm$ for $f1$ and $8.00 cm$ for $p1$ in above expression.

$18.00 cm+1q1=15.00 cmq1=13.33 cm$

Thus the image distance from the lens is $13.33 cm$ .

The expression for the object distance for mirror is,

$p2=d−q1$

Here,

$d$ is the distance between the lens and mirror.

Substitute $13.33 cm$ for $q1$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−13.33 cm=6.67 cm$

Thus the object distance for mirror is $6.67 cm$ .

The expression for the mirror equation is,

$1p2+1q2=1f2$

Here,

$q2$ is the image distance for mirror.

$p2$ is the object distance for mirror.

$f2$ is the focal length for mirror.

Substitute $4.00 cm$ for $f2$ and $6.67 cm$ for $p2$ in above expression.

$16.67 cm+1q2=14.00 cmq2=10.0 cm$

Thus the image distance from the mirror is $10.0 cm$ .

The expression for the object distance for lens is,

$p3=d−q2$

Substitute $10.0 cm$ for $q2$ and $20.0 cm$ for $d$ in above expression.

$p2=20.0 cm−10.0 cm=10 cm$

Thus the object distance for lens is $10.0 cm$ .

The expression for the lens equation is,

$1p3+1q3=1f3$

Here,

$q3$ is the image distance for lens.

$p3$ is the object distance for lens.

$f3$ is the focal length for lens.

Substitute $5.00 cm$ for $f3$ and $10.0 cm$ for $p3$ in above expression.

$110.0 cm+1q3=15.00 cmq3=10.0 cm$

Conclusion:

Therefore the position of the final image is $10.0 cm$ to the left of the lens.

Answer 22

(c)

Expert Solution

To determine

The magnification as seen by the eye in the figure.

Answer to Problem 36.95CP

The magnification as seen by the eye in the figure is $−2.50$ .

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The overall magnification factor is,

$M=(−q1p1)(−q2p2)(−q3p3)$

Substitute $13.33 cm$ for $q1$ , $8.0 cm$ for $p1$ , $10.0 cm$ for $q2$ , $6.67 cm$ for $p2$ , $10.0 cm$ for $q3$ and $10.0 cm$ for $p3$ in above expression.

$M=(−13.33 cm8.0 cm)(−10 cm6.67 cm)(−10 cm10 cm)=−2.50$

Conclusion:

Therefore the magnification as seen by the eye in the figure is $−2.50$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The magnification as seen by the eye in the figure.

Answer 27

Answer to Problem 36.95CP

The magnification as seen by the eye in the figure is $−2.50$ .

Answer 28

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

The overall magnification factor is,

$M=(−q1p1)(−q2p2)(−q3p3)$

Substitute $13.33 cm$ for $q1$ , $8.0 cm$ for $p1$ , $10.0 cm$ for $q2$ , $6.67 cm$ for $p2$ , $10.0 cm$ for $q3$ and $10.0 cm$ for $p3$ in above expression.

$M=(−13.33 cm8.0 cm)(−10 cm6.67 cm)(−10 cm10 cm)=−2.50$

Conclusion:

Therefore the magnification as seen by the eye in the figure is $−2.50$ .

Answer 29

(d)

Expert Solution

To determine

Whether the final image is inverted or upright.

Answer to Problem 36.95CP

The final image is inverted.

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

From part (c) of the question the value of the magnification factor is,

$M=−2.50$

It is known that the image will be upright when the magnification factor is positive and the image will be inverted when the magnification factor is negative.

Since the magnification factor is negative so the image will be upright.

Conclusion:

Therefore the final image is inverted.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

Whether the final image is inverted or upright.

Answer 34

Answer to Problem 36.95CP

The final image is inverted.

Answer 35

Explanation of Solution

Given data: The radii of curvature for the converging Len’s surfaces are $9.00 cm$ and $11.0 cm$ , the radius of curvature of the mirror is $8.00 cm$ and the distance between the focal points $F1$ and $F2$ is $5.00 cm$ , the distance between the lens and mirror is $20.0 cm$ and the object distance for the lens is $8.00 cm$ .

From part (c) of the question the value of the magnification factor is,

$M=−2.50$

It is known that the image will be upright when the magnification factor is positive and the image will be inverted when the magnification factor is negative.

Since the magnification factor is negative so the image will be upright.

Conclusion:

Therefore the final image is inverted.